如何在f＃中解包列表中的联合值

Question

You know that to unwrap a value of a single union type you have to do this:

type Foo = Foo of int*string

let processFoo foo =
    let (Foo (t1,t2)) = foo
    printfn "%A %A" t1 t2 

but my question is: if there a way to do that for lists ?:

let processFooList (foolist:Foo list )  =
    let ??? = foolist // how to get a int*string list
    ...

thanks.

Answer 1

最好的方法是使用一个与List.map结合的函数

let processFooList (foolist:Foo list )  = foolist |> List.map (function |Foo(t1,t2)->t1,t2)

Answer 2

There's no predefined active pattern for converting the list from Foo to int * string , but you could combine the Named Pattern §7.2 (deconstruct single case union) with the projection into your own single case Active Pattern §7.2.3 .

let asTuple (Foo(t1, t2)) = t1, t2      // extract tuple from single Foo
let (|FooList|) =  List.map asTuple     // apply to list

Use as function argument:

let processFooList (FooList fooList) =  // now you can extract tuples from Foo list
    ...                                 // fooList is an (int * string) list

Use in let-binding:

let (FooList fooList) = 
    [ Foo(1, "a"); Foo(2, "b") ]
printfn "%A" fooList                    // prints [(1, "a"); (2, "b")]

Answer 3

Distilling/summarising/restating/reposting the other two answers, your cited line:

let ??? = foolist // how to get a int*string list

Can become:

let ``???`` = foolist |> List.map (function |Foo(x,y) -> x,y)

If you're writing a transformation, you can do the matching in the params having defined an Active Pattern using either of the following:

let (|FooList|) = List.map <| fun (Foo(t1, t2)) -> t1,t2
let (|FooList|) = List.map <| function |Foo(t1, t2) -> t1,t2

which can then be consumed as follows:

let processFooList (fooList:Foo list )  =
    // do something with fooList